Deceptively simple mass-spring problem?

Got it, I think.

As $m_1$ drops, gravitational potential energy is converted to spring potential energy:

$$m_1gy=\frac12 k\ell^2$$ Where $y=\ell$ because the string doesn't stretch.

The maximum drop is: $$y_\textrm{max}=\frac{2m_1g}{k}$$ Maximum tension in the spring then is: $$T_\textrm{max}=ky_\textrm{max}=2m_1g$$ For $m_2$ to move: \begin{align}2m_1g &\gt 2m_2g\sin\theta\\ \implies m_1 &\gt m_2\sin\theta\end{align}

If that condition isn't met, $m_2$ won't move and $m_1$ will start moving upwards again and enter an oscillatory motion.

If $m_1$ is too small. The mass fall down and bounces back up. The force on the masses is highest at the lowest point of the bounce. A mass that is just large enough will also bounce back but at the lowest point of the bounce the force will just be large enough to move $m_2$. So we have to find an expression for the force at the lowest point.

Using conservation of energy you can find how far the mass falls by equating the loss in gravitational energy with the gain in energy stored in the spring. From there I guess you should be able to figure it out

You have derived an equation which predicts the minimum force $F_{\text{min}}$ required to get the block $m_2$ moving up the slope.

$$F_{\text{min}} = \mu m_2g\cos \theta+m_2g\sin \theta$$

Until the tension in the string is equal to that value of force the block $m_2$ will not move so the spring mass $m_1$ system can be thought of a a standard spring-mass system with block $m_2$ being the rigid support.

The static equilibrium position of the spring-mass $m_1$ system is when $m_1 g - k l =0$ where $k$ is the spring constant and $l$ is the static extension of the spring.
When mass $m_1$ is released the magnitude of the force at both ends of the spring increases as the spring stretches.
However the mass $m_1$ overshoots the static extension position and continues on downwards until the extension of the spring is a maximum, $= 2l$, and hence the force on each end of the spring is $2k_l = 2 m_1g$ where the mass $m_1$ stops.
This is the maximum force exerted by the spring on block $m_2$ via the string and is your required $F_{\text{min}}$.

The key to this is the fact that the spring is at its rest length, and m1 is released. The question is, what happens then? Note: your answer will be different than it would be if the system was initially in equilibrium.

If we assume that M1 is small enough not to dislodge M2 we can handle it as a spring on rigid support as mentioned above.
If it is heavy enough to move the M2 then the system acts in two configurations:
1- $-\omega^2.M_1g<M_2.\mu.g. cos(\theta) $ note we still don't
consider the effect of inertia of M2.
$\omega = (k/M_1)^.5$

2- when $-\omega^2.M_1g>M_2.\mu.g. cos(\theta) $

$\mu m_2g\cos \theta+m_2g\sin \theta - M_1.g -K.X=0 $

From here on we can use Duhamel integration to drive the equation which I am not going to do! because it takes time.

**Added a sketch **
Hopefully this crude sketch will clarify a bit. Please note the segments on the bottom graph indicating the movement of CG are not straight lines Also the movement of CG may be to reverse direction depending on friction and mass of M2.

Like you reasoned, the following is true about friction:

$$\mu_s=\tan\theta$$

I'd start with drawing a free body diagram of $m_1$ and coming up with the following equation, assuming that $m_2$ never moves:

$$m_1g - kx = m_1\frac{d^2x}{dt^2}$$

Then consider $m_2$ assuming it doesn't move:

$$kx=m_2g{\sin\theta} + \mu_sm_2g\cos\theta$$

Or, at maximum spring tension:

$$kx=2m_2g{\sin\theta}$$

But we can remember that:

$$\frac{d^2x}{dt^2} = v\frac{dv}{dx}$$

We can substitute that back into the differential equation:

$$v\frac{dv}{dx} = g - \frac{k}{m_1}x$$

Then integrate:

$$\int v{dv} = \int gdx - \frac{k}{m_1}xdx$$

This yields:

$$\frac{v^2}{2} = gx - \frac{1}{2} \frac{k}{m_1}x^2$$

At maximum tension, $v$ equals $0$, which yields after substituting in the maximum tension $kx$:

$$m_1 = m_2{\sin\theta}$$

I ended up with the same solution as before, only this time, I integrated properly. It's funny how the incorrect procedure yielded the same result, but it's explained by the following fact:

$$\frac{d^2x}{dt^2} = v\frac{dv}{dx}$$

Let me try to do this without using any formulas. First consider what would happen if $m_2$ were glued to the incline. Then we would have a simple harmonic oscillator consisting of $m_1$ suspended from the spring $k$ (clearly it does not matter on which side of the pulley the spring is). In this situation the average tension force (over a full oscillation) of the spring must equal the weight of $m_1$, by conservation of momentum. The initial position corresponds to the topmost position of the oscillation, and it is given that the spring has zero tension in this position. Then to get the predicted average over an oscillation, the tension must be twice the weight of $m_1$ halfway the oscillation, when $m_2$ is at its lowest position.

Since this is the largest tension that occurs during the cycle, the situation where $m_2$ is not glued to the surface will change as soon as this maximal tension is sufficient to make it start to slide upward. It follows that your original static computation overestimates the required mass of $m_1$ by a factor$~2$.

This is not really a steady problem.

There is overshoot from the spring, which means the maximum and steady force exerted by the mass is not, in general, equal. This difference means that the kinetic coefficient of friction is important, since movement may start at the maximum force exerted by the mass-spring. This is the conceptual issue with you're problem statement.

If the problem were truly steady, or damped (with, e.g. A dash pot system), then the spring has no impact on your final result.